Problem: Let $f(x)=\dfrac1xe^x$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{x^2} e^x$ (Choice B) B $\dfrac{1}{x} e^x$ (Choice C) C $-\dfrac{1}{x^2}\left(\dfrac1x+e^x\right)$ (Choice D) D $e^x\left(\dfrac1x-\dfrac{1}{x^2}\right)$
Solution: $f(x)$ is the product of two, more basic, expressions: $\dfrac1x$ and $e^x$. Therefore, the derivative of $f$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac1xe^x\right) \\\\ &=\dfrac{d}{dx}\left(\dfrac1x\right)e^x+\dfrac1x\dfrac{d}{dx}(e^x)&&\gray{\text{The product rule}} \\\\ &=\dfrac{d}{dx}\left(x^{-1}\right)e^x+\dfrac1x\dfrac{d}{dx}(e^x)&&\gray{\text{Write }\dfrac1x \text{as a power}} \\\\ &=-1x^{-2}\cdot e^x+\dfrac1x\cdot e^x&&\gray{\text{Differentiate }x^{-1}\text{ and }e^x} \\\\ &=-\dfrac{1}{x^{2}}\cdot e^x+\dfrac1x\cdot e^x&&\gray{\text{Rewrite with positive exponents}} \\\\ &=e^x\left(\dfrac1x-\dfrac{1}{x^2}\right)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $f'(x)=e^x\left(\dfrac1x-\dfrac{1}{x^2}\right)$